Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello guys, I am taking a look at the "Gaussian Process Latent Variable Models" (GP-LVM) from Lawrence's paper (LINK) and because of my lack of experience in Bayesian modeling I am having some trouble with his derivation; I hope you can help :)


Basic formulation

Following the paper, $\mathbf{Y}=[\mathbf{y}_1, ..., \mathbf{y}_n]^T$ is the observations matrix and similarly $\mathbf{X}$ is the matrix of latent variables. The distributions are gaussian and defined as

$$p(\mathbf{y}_n | \mathbf{x}_n, \mathbf{W}, \beta) = N(\mathbf{y}_n | \mathbf{W}\mathbf{x}_n, \beta^{-1}\mathbf{I}) \quad \quad p(\mathbf{x}_n)=N(\mathbf{x}_n|0,\mathbf{I})$$

Assuming i.i.d. his the given relationship in between latent space and observations is:

$$p(\mathbf{Y} | \mathbf{W},\beta) = \prod_{n=1}^N p(\mathbf{Y}_n | \mathbf{W})$$

What I am interested in understanding is how he obtains Eq.(1), that is, the expression for $p(\mathbf{Y}|\mathbf{X},\beta)$:

$$p(\mathbf{Y}|\mathbf{X},\beta) = \text{(scale)} \: \text{exp} \left( -\frac{1}{2} tr( \mathbf{K}^{-1} \mathbf{Y} \mathbf{Y}^T) \right), \quad \mathbf{K}=\alpha \mathbf{XX}^T + \beta^{-1}\mathbf{I} $$

Note that to achieve this he assumes that the linear map $\mathbf{W}$ is distributed as $$p(W)=\prod_{i=1}^{D} N(\mathbf{w}_i | 0, \alpha^{-1}\mathbf{I})$$


Begin of self-explanation

What I believe he's doing is first removing the conditional relationship w.r.t. $\mathbf{W}$ by:

$$p(\mathbf{Y},\mathbf{W}|\beta) = p(\mathbf{Y}|\mathbf{W},\beta) p(\mathbf{W})$$

Then, he marginalizes w.r.t. $\mathbf{W}$ to get rid of the dependence from the linear mapping:

$$p(\mathbf{Y} | \mathbf{X}, \beta) = \int_{\mathbf{W}} p(\mathbf{Y},\mathbf{W}|\beta) $$

Now, if what I have above is correct, my simplified question becomes:

$$ \mathbf{\text{how do you integrate w.r.t. a matrix?}} $$

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.